Physics, asked by Abhisheksagar49, 11 months ago

A body is rolling without slipping on horizontal surface
If it has its rotational kinetic energy equal to the
translational kinetic energy, then the body is :​

Answers

Answered by sanchitdubey388
10

Answer:

whereas the rotational kinetic energy is

( 2 )

KErot = (1/2)Iω2

In this last equation ω is the angular velocity in radians/sec, and I is the object's moment of inertia. For objects with simple circular symmetry (e.g. spheres and cylinders) about the rotational axis, I may be written in the form:

( 3 )

I = kmr2

where m is the mass of the object and r is its radius. The geometric factor k is a constant which depends on the shape of the object:

k = 2/5 = 0.4 for a uniform solid sphere,

k = 1/2 = 0.5 for a uniform disk or solid cylinder,

k = 1 for a hoop or hollow cylinder.

If the object rolls without slipping, then the object's linear velocity and angular speed are related by

v = rω.

Substituting equation 3 and the expression for v into equation 2, we obtain:

Answered by soniatiwari214
0

Answer:

On a flat and horizontal surface, a body is rolling without slipping. If it has its rotational kinetic energy equal to the translational kinetic energy, then the body is a ring.

Explanation:

  • The effort necessary to accelerate a rigid body from rest to a certain velocity is known as the object's translational kinetic energy. Motion from one location to another that follows a straight line is referred to as translational motion.
  • An object's rotation generates kinetic energy, also known as angular kinetic energy, which is one of the components that determine the object's overall kinetic energy.
  • Mathematically, translational and rotational kinetic energy is expressed as Kt = 1/2 mv² and Kr = 1/2 Iω²
  • Equating both the equations we get,
  • 1/2 mv² = 1/2 Iω² ---- equation 1
  • Since the object is rolling without slipping thus, a relation is followed - v = ωr  ---- equation 2
  • Substitute equation 2 in 1, we get - 1/2 mω²r² = 1/2 Iω²
  • Therefore, I = mr²
  • The above moment of inertia of the ring corresponds to the moment of inertia of the ring.

Thus, a body is rolling without slipping on a horizontal surface. If it has its rotational kinetic energy equal to the translational kinetic energy, then the body is a ring.

#SPJ3

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